Question
Download Solution PDFThe steady flow energy equation:
Q = m(h2 – h1) is applicable for?Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Steady Flow Energy Equation (S.F.E.E.)
\(m\left( {{h_1} + \;\frac{{v_1^2}}{2} + g{z_1}} \right) + \dot{Q} = m\left( {{h_2} + \frac{{v_2^2}}{2} + g{z_2}} \right) + \dot{W}\)
Now,
Boiler:
- Ẇ = 0, no work is done by the boiler,
- Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)
∴ Q̇ = ṁ(h2 - h1)
Nozzle:
- Q̇ = 0, as the nozzle is perfectly insulated.
- Ẇ = 0, no work is done by the nozzle, v1 <<< v2 and potential energy change is neglected
∴ \({v_2} = \sqrt {2\left( {{h_1} - {h_2}} \right)}\)
Turbine:
- Q = 0, for the adiabatic or perfectly insulated turbine,
- Change in kinetic and potential energy is neglected
∴ Ẇ = ṁ(h1 – h2)
Compressor:
- Q̇ = 0, for the adiabatic or perfectly insulated turbine,
- Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0)
∴ Ẇ = ṁ(h2 - h1)
Points to remember:
Device |
S.F.E.E |
Nozzle |
\({v_2} = \sqrt {2\;\left( {{h_1} - {h_2}} \right)}\) |
Turbine |
Ẇ = ṁ(h1 – h2) |
Compressor |
Ẇ = ṁ(h2 - h1) |
Boiler |
Q = ṁ(h2 - h1) |
Last updated on May 28, 2025
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